Two corners of an isosceles triangle are at #(8 ,1 )# and #(1 ,4 )#. If the triangle's area is #15 #, what are the lengths of the triangle's sides?

Question

1137 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-30T11:16:33

Lengths of triangle's sides are #7.62 , 5.48 , 5.48 # unit

Base of the isosceles triangle is #b=sqrt((x_1-x_2)^2+(y_1-y_2)^2) =sqrt((8-1)^2+(1-4)^2)=sqrt58=7.62(2dp)#

Area of the triangle is #A_t =1/2*b*h or h= (2*A_t)/b = (2*15)/7.62=30/7.62=3.94(2dp)#. where #h# is altitude.

Legs are #l= sqrt (h^2+(b/2)^2) = sqrt (3.94^2+(7.62/2)^2)=sqrt (30.04)=5.48(2dp)#

Lengths of triangle's sides are #7.62 , 5.48 , 5.48 (2dp)#unit[Ans]